Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Mixed Review Exercises - Page 362: 6


$5y^2 \left( 3y+4 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ 15y^3+20y^2 .$ Divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 15,20 \}$ is $ 5 $ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ y^3,y^2 \}$ is $ y^2 .$ Hence, the entire expression has $GCF= 5y^2 .$ Factoring the $GCF= 5y^2 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 5y^2 \left( \dfrac{15y^3}{5y^2}+\dfrac{20y^2}{5y^2} \right) \\\\= 5y^2 \left( 3y+4 \right) .\end{array}
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